Write the function $\log(1+x), x\in(-1,1]$ by expanding it into an infinite series by application of Taylor's Theorem.
I tried solving this as follows:
We know that,
(Taylor's Theorem in Cauchy's form of remainder)
If $f(x)$ is a function such that,
$f^{n-1}(x)$ is continuous on $[a,a+h]$
$f^n(x)$ exists in $(a,a+h)$
Then, $f(a+h)=f(a)+hf'(a)+h^2\frac{f^{(2)}(a)}{2!}+\cdots +h^{n-1}\frac{f^{((n-1))}(a)}{(n-1)!}+h^n\frac{f^{(n)}(a+\theta h)}{n!},$ where $\theta\in (0,1).$
Now, if $h=x$ and $a=0$ we get, Maclaurin's Theorem, i.e
If $f(x)$ is a function such that,
$f^{n-1}(x)$ is continuous on $[0,x]$
$f^n(x)$ exists in $(0,x)$
Then, $f(x)=f(0)+xf'(0)+x^2\frac{f^{(2)}(0)}{2!}+\cdots +x^{n-1}\frac{f^{((n-1))}(0)}{(n-1)!}+x^n\frac{f^{(n)}(\theta x)(1-\theta)^{n-1}}{(n-1)!}\tag 1,$ where $\theta\in (0,1).$
Here, $f(x)=\log(1+x)$ where $x\in(-1,1].$
We have, $f^{(n)}(x)=\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}.$
So, $f^{(n)}(\theta x)=\frac{(-1)^{n-1}(n-1)!}{(1+\theta x)^n}$ and $f^{(n)}(0)=(-1)^{n-1}(n-1)!.$
We assume the Lagrange's form of remiander $R_n=x^n\frac{f^{(n)}(\theta x)(1-\theta)^{n-1}}{(n-1)!}$ in $(1)$
We let $(s_n)$ be a sequence of partial sums where, $$s_n=f(0)+xf'(0)+x^2\frac{f^{(2)}(0)}{2!}+\cdots +x^{n-1}\frac{f^{((n-1))}(0)}{(n-1)!}+x^n\frac{f^{(n)}( 0)}{n!}=0+1-\frac{x^2}{2}+\cdots+\frac{x^n}{n!}(-1)^{n-1}(n-1)!=0+1-\frac{x^2}{2}+\cdots+(-1)^{n-1}\frac{x^n}{n}.$$
We note that if $(s_n)$ is the sequence of partial sums of the following sequence, $(u_n),$ where $u_n=(-1)^{n-1}\frac{x^n}{n}.$
This means $u_{n+1}=(-1)^{n}\frac{x^{n+1}}{n+1}$ and so, $\lim (\frac{u_{n+1}}{u_n})=-\frac{nx}{n+1}=\lim-x(1-\frac{1}{n+1})=-x.$
By D'Alambert's Ratio Test, we have, $x\gt -1\implies -x\lt 1\implies $ $(s_n)$ is convergent.
Now, $$\lim R_n=\lim x^n\frac{f^{(n)}(\theta x)(1-\theta)^{n-1}}{(n-1)!}=\lim x^n\frac{(-1)^{n-1}(1-\theta)^{n-1}}{(1+\theta x)^n}$$ needs to be evaluated. We have, if $x\in(-1,1)$ then, $\lim x^n=0$ and also, the sequence $y_n=(-1)^n$ is bounded.
Now, $\frac{1}{1+\theta x}$ is finite.
But the problem, comes with, $(\frac{1-\theta}{1+\theta x})^n$. I really cant, say anything about this term.
In the book, a hint is given, saying that, $$\big|\frac{1-\theta}{1+\theta x}\big |\lt 1.$$ But I am been able to prove this inequality.
I need some help, to proceed from here.
The derivation of logarithmic series using Taylor's theorem has some intricacies which I deal here.
By the Taylor's theorem we have $$f(x) =\sum_{k=0}^n\frac{x^k}{k!}f^{(k)}(0)+R_{n+1}(x)$$ where $R_{n+1}(x)$ can be expressed in many forms to allow for its estimation and thereby evaluate its limit as $n\to\infty $.
If $f(x) =\log(1+x)$ then $$f^{(k)} (x) =(-1)^{k-1}(k-1)!(1+x)^{-k}$$ Let $0\leq x\leq 1$ and we use the Lagrange form of remainder $$R_{n+1}(x)=\frac{x^{n+1}}{(n+1)!}f^{(n+1)}(\theta_{n+1} x) $$ where $\theta_{n+1}\in(0,1)$ depends on $x$ as well as $n$. We then have $$|R_{n+1}(x)|=\frac{x^{n+1}}{(n+1)(1+\theta_{n+1}x)^{n}}\leq \frac{x^{n+1}}{n+1}$$ And thus $R_{n+1}(x)\to 0 $ for all $x\in[0,1]$ as $n\to\infty$.
For $x\in(-1,0)$ the estimate via Lagrange form of remainder doesn't really help in finding its limit. And then we use the Cauchy form of remainder given by $$R_{n+1}(x)=\frac{x^{n+1}(1-\theta_{n+1})^n}{n!}f^{(n+1)}(\theta_{n+1}x)$$ and we have $$|R_{n+1}(x)|=\frac{|x|^{n+1}}{1+\theta_{n+1}x}\left(\frac{1-\theta_{n+1}}{1+\theta_{n+1}x}\right)^{n}$$ The fraction in large parentheses is positive (as both numerator and denominator are positive) and further it is less than $1$ because $$-1<x\implies 1-\theta_{n+1}<1+\theta_{n+1}x$$ Hence $$|R_{n+1}(x)|<\frac{|x|^{n+1}}{1+\theta_{n+1}x}<\frac{|x|^{n+1}}{1+x}$$ so that $R_{n+1}(x)\to 0$.
This completes the analysis of remainder in Taylor's theorem and we have $$\log(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{k}$$ for all $x\in(-1,1]$.